Companion to the iron trade etc., a series of comprehensive tables1851 |
Common terms and phrases
14 inch 20 feet ALLOYS antimony Bar-Iron BD BD beam will sustain bearing in feet bismuth Blast brass breadth in inches Cast Iron ciphers circular Segments Circum circumference co co co column copper cube root cubic inches cylinder decimal Deflection in inches depth in inches diameter in inches divide the product ends Example.-Required Example.-What exterior diameter Fahrenheit feet long find the area find the load Flanges gallon Hoops inch thick inches broad Inches Inches Inches Iron-Merchant length in feet length of bearing load in lbs Messrs Metal multiply ounces perpendicular Plate Pounds PROBLEM quotient radius rectangular rivets RULE RULE.-Multiply RULE.-Multiply the length s. d. Cwt s. d. qr segments side solid content square inch square root subtract supported in lbs Table tons tube weight in lbs whole numbers Wire Gauge WROUGHT IRON yards zinc دو ਨੂੰ ᎧᎧ
Popular passages
Page 233 - 3 0 I 3 6 4 0 4 6 5 0 5 6 6 0 6 6 7 0 7 6 8 0 Bursting pressure — equivalent to the ultimate strength of the riveted joint — -as deduced from experiment« 34,000 Ibs. to the square inch. 450 Ibs. Thickness of the Plates in decimal parts of an inch. •250 •291 •333
Page 175 - rule of last Problem. Find also the area of the triangle formed by the chord of the segment and the two radii of the sector; then add these together for the answer, when the segment is greater than a semicircle ; or subtract them, •when it is less than a semicircle : as is evident. RULE
Page 177 - RULE.—Multiply the perimeter of the base by the slant height, or length of the side, and half the product will be the surface of the sides, or the sum of the areas of all the triangles which form it. To which add the area of the end or base, if required.
Page 187 - apply to this Problem, only changing the divisors, or halving the quotients. PROBLEM IV. To find the dimensions, when a beam is fixed at one end and loaded at the other, or when it is supported at the middle and loaded at both ends.
Page 178 - RULE.—Add into one sum the areas of the two ends, and the mean proportional between them ; and take one-third of that sum for a mean area; which, being multiplied by the perpendicular height or length of the frustum, will give its content.
Page 180 - RULE.—To twice the length of the base add the length of the edge ; multiply the sum by the breadth of the base, and by the perpendicular height from the base, and one-sixth of the product is the solid
Page 177 - RULE.—Multiply the perimeter of one end of the Prism, by the length of the solid, and the product will be the surface of all its sides. To which add also the area of the two ends of the Prism, when required. Or, compute the areas of all the sides and ends separately, and add them all together.
Page 179 - three times the square of the radius of the Segment's base, add the square of its height ; then multiply the sum by the height, and the product by '5236 for the content.
Page 184 - by the square of the depth in inches, and four times the breadth; divide that product by the length in inches, and the quotient will be the weight in Ibs. Example I.—What weight will break a beam of English Oak 7 inches broad, 9 inches deep, and 30 feet between the props ? 1672 x 81 x 28
Page 179 - three times the diameter of the Sphere take double the height of the Segment; then multiply the remainder by the square of the height, and the product by the decimal -5236 for the content. RULE